Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
double dp[3000][3000];
int n;
double p[3000];
double ans;

int main(){
	cin>>n;
	for(int i=1;i<=n;i++) cin>>p[i];
	dp[0][0]=1;
	for(int i=1;i<=n;i++){
		for(int j=0;j<=i;j++){
			dp[i][j]+=dp[i-1][j]*(1.0-p[i]);
			if(j>0) dp[i][j]+=dp[i-1][j-1]*p[i];
		}
	}
	for(int i=0;i<=n;i++){
		if(i>n-i) ans+=dp[n][i];
	}
	printf("%.10lf\n",ans);
	
	return 0;
}